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exis [7]
3 years ago
10

How does math relate to electrons and ion formation?

Chemistry
2 answers:
Elenna [48]3 years ago
7 0
In order to find the NET nuclear charge from an atom's valence electron to the proton nucleus, you need to do some simple math in order to find the charge.

You would also need to apply math when you need to figure out what the charge on an ion is by either adding or taking away electrons depending on whether it is an anion or cation.
AlladinOne [14]3 years ago
5 0

Answer and explanation;

Ions are formed when an atom looses or gains electrons.

-Ionic charges are created when there is an imbalance of protons and electrons on an atom. Since we can not change the number of protons on the nucleus of an atom, a positively charged ion will be created when there are fewer electrons than protons (in other words when an atom looses electron). A negatively charged ion is formed or created when there are more electrons than protons in an atom (when an atom gains electrons).

-When electron are added, you are adding more negative charge and thus, results in the creation of a negative ion (anion), and when electrons are being lost by an atom, one is removing negative charge, resulting in creation of a positive ion (cation).

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3 years ago
Which of the following has the strongest buffering capacity? A. H2O B. 0.1 M HCl C. 0.1 M carbonic/bicarbonate (H2CO3/HCO3-) at
enyata [817]

Explanation:

(A)   As we know that carbonic acid (H_{2}CO_{3}) and Sodium bicarbonate (NaHCO_{3}) forms an acidic buffer.

Therefore, pH of an acidic buffer is given by Hendeerson-Hasselbalch equation as follows.

               pH = pK_{a} + log(\frac{[Salt]}{[Acid]}) ........... (1)

So mathematically,  if [Salt] = [Acid]  then \frac{[Salt]}{[Acid]} = 1 .

And,  log (\frac{[Salt]}{[Acid]}) = 0

Therefore, equation (1) gives us the following.

         pH = pK_{a} (when acid and salt are equal in concentration)

Hence, pK_{a} of H_{2}CO_{3} (carbonic acid) is 6.35.

And, with this we have following results.

In (A) and (D) we have the case \frac{[NaHCO_{3}]}{[H_{2}CO_{3}]}[/tex] i.e. [Salt] = [Acid].

Hence, for the cases pH = pK_{a} = 6.35.

(B)    [NaHCO_{3}] = 0.045 M and,  [H_{2}CO_{3}] = 0.45 M

Hence,   pH = 6.35 + log([NaHCO_{3}][[H_{2}CO_{3}])

                     = 6.35 + log(\frac{0.045}{0.45})

                     = 6.35 + (-1)

                     = 5.35

Therefore, it means that this buffer will be most suitable buffer as it has pH on acidic side and addition of slight excess base will not affect much of its pH value.

(C)    [NaHCO_{3}] = 0.45 M [H_{2}CO_{3}]

                          = 0.045 M

So,       pH = 6.35 + log(\frac{[NaHCO_{3}]}{[H_{2}CO_{3}]})

                  = 6.35 + log(\frac{0.45}{0.045})

                  = 6.35 + (+1)

                 = 7.35

This means that pH on Basic side makes it no more acidic buffer.

5 0
3 years ago
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